Convert the point $(2 \sqrt{3}, 6, -4)$ in rectangular coordinates to spherical coordinates.  Enter your answer in the form $(\rho,\theta,\phi),$ where $\rho > 0,$ $0 \le \theta < 2 \pi,$ and $0 \le \phi \le \pi.$
We have that $\rho = \sqrt{(2 \sqrt{3})^2 + 6^2 + (-4)^2} = 8.$  We want $\phi$ to satisfy
\[-4 = 8 \cos \phi,\]so $\phi = \frac{2 \pi}{3}.$

We want $\theta$ to satisfy
\begin{align*}
2 \sqrt{3} &= 8 \sin \frac{2 \pi}{3} \cos \theta, \\
6 &= 8 \sin \frac{2 \pi}{3} \sin \theta.
\end{align*}Thus, $\theta = \frac{\pi}{3},$ so the spherical coordinates are $\boxed{\left( 8, \frac{\pi}{3}, \frac{2 \pi}{3} \right)}.$